Let a → , b → , c → be three non-coplanar vectors such that a → × b → = 4…

Taking cross product of $\overrightarrow{c}$ both side we get,

$\overrightarrow{a}\times \overrightarrow{b}=4\overrightarrow{c}\Rightarrow \left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}=0$

$\Rightarrow \left(\overrightarrow{a}·\overrightarrow{c}\right).\overrightarrow{b}-\left(\overrightarrow{b}·\overrightarrow{c}\right).\overrightarrow{a}=0$

$\Rightarrow \left(\overrightarrow{a}·\overrightarrow{c}\right).\overrightarrow{b}=\left(\overrightarrow{b}·\overrightarrow{c}\right).\overrightarrow{a}$

Since vectors are non-coplanar so,

$\Rightarrow \left(\overrightarrow{a}·\overrightarrow{c}\right)=\left(\overrightarrow{b}·\overrightarrow{c}\right)=0 ......\left(1\right)$

Similarly for $\overrightarrow{b}\times \overrightarrow{c}=9\overrightarrow{a}\Rightarrow \overrightarrow{a}·\overrightarrow{b}=0=\overrightarrow{a}·\overrightarrow{c} .....\left(2\right)$

So, from equation $\left(1\right) \& \left(2\right)$ we can say that  $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually $\perp$ set of vectors.

Now taking modulus both side of $\overrightarrow{a}\times \overrightarrow{b}=4\overrightarrow{c}, \overrightarrow{b}\times \overrightarrow{c}=9\overrightarrow{a} \& \overrightarrow{c}\times \overrightarrow{a}=\alpha \overrightarrow{b}$ we get,

$\Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|=4\left|\overrightarrow{c}\right|,\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=9\left|\overrightarrow{a}\right| \& \left|\overrightarrow{c}\right|\left|\overrightarrow{a}\right|=\alpha \left|\overrightarrow{b}\right|$

$\Rightarrow \frac{\left|\overrightarrow{a}\right|}{\left|\overrightarrow{c}\right|}=\frac{4}{9}\frac{\left|\overrightarrow{c}\right|}{\left|\overrightarrow{a}\right|}$

$\Rightarrow \frac{\left|\overrightarrow{\mathrm{c}}\right|}{\left|\overrightarrow{\mathrm{a}}\right|}=\frac{3}{2}$

$\therefore$  If $\left|\mathrm{a}\right|=\lambda ,\left|\mathrm{c}\right|=\frac{3\lambda }{2} \& \left|\mathrm{b}\right|=6$

Now $\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|=36$

$\Rightarrow \lambda +\frac{3}{2}\lambda +6=36$

$\Rightarrow \lambda =12$

Now given $\left|\overrightarrow{c}\right|\left|\overrightarrow{a}\right|=\alpha \left|\overrightarrow{b}\right|$

$\Rightarrow \alpha =\frac{\left|\overrightarrow{c}\right|\left|\overrightarrow{a}\right|}{\left|\overrightarrow{b}\right|}=\frac{3\times 12}{2}\times \frac{12}{6}$

$\Rightarrow \alpha =36$

 

Note this question was bonus in jee main 27th july 2022 shift 2, so we have some modification to the originial question.